Question

Let $\alpha$ and $\beta$ be the distinct roots of $a{x}^2+bx+c=0$, then $\underset{x\to \alpha }{lim}\frac{1-\cos (a{x}^2+bx+c)}{(x-\alpha {)}^2}$ is equal to

• A. $0$
• B. $\frac{1}{2}{a}^2(\alpha -\beta {)}^2$
• C. $\frac{1}{2}(\alpha -\beta {)}^2$
• D. $-\frac{1}{2}{a}^2(\alpha -\beta {)}^2$

Answer 1

The correct option is B $\frac{1}{2}{a}^2(\alpha -\beta {)}^2$

Consider, $\underset{x\to \alpha }{lim}\left(\frac{1-\cos (a{x}^2+bx+c)}{{(x-\alpha )}^2}\right)$

$=\underset{x\to \alpha }{lim}\left(\frac{2{\sin }^2\left[\frac{(a{x}^2+bx+c)}{2}\right]}{{(x-\alpha )}^2}\right)$ ... $[\because \cos 2x=1-2{\sin }^{2}x]$

$=\underset{x\to \alpha }{lim}2\left(\frac{{\sin }^2\left[a\frac{(x-\alpha )(x-\beta )}{2}\right]}{{\left(a\frac{(x-\alpha )(x-\beta )}{2}\right)}^2}\right)\times \frac{1}{4}{a^2(x-\beta )}^2$

$=2\times \frac{1}{4}{a}^2{(\alpha -\beta )}^2$

$=\frac{1}{2}{a}^2{(\alpha -\beta )}^2$